Pressure: Fundamental Principles and Applications in Fluids and Gases

Defining Pressure

Pressure (\(P\)) quantifies force distribution over a surface area:
\[
P = \frac{F_{\perp}}{A}
\]
Where:

• \(F_{\perp}\): Normal force component (N)
• \(A\): Contact area (m²)
• 1 Pa = 1 N/m²

Fluid Pressure

Hydrostatic Pressure

\[
P = P_0 + \rho gh
\]
Where:

• \(P_0\): Atmospheric pressure (101.3 kPa)
• \(\rho\): Fluid density (kg/m³)
• \(g\): Gravitational acceleration (9.81 m/s²)
• \(h\): Depth (m)

Pascal’s Principle

Pressure transmission enables hydraulic force multiplication:
\[
\frac{F_1}{A_1} = \frac{F_2}{A_2}
\]

Gas Pressure

Ideal Gas Law

\[
PV = nRT
\]
Constants:

• \(R = 8.314 \, \text{J/(mol·K)}\)
• STP: 273.15 K, 101.325 kPa

Practical Applications

Engineering Systems

• Hydraulic lifts: Achieve 100:1 force amplification
• Scuba diving: Pressure increases ~101 kPa per 10m depth

Meteorology

• Barometric pressure: 98-104 kPa typical range
• Weather fronts: ~5 kPa pressure differences

Worked Example

Water Pressure at 5m Depth:
\[
P = 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 5 \, \text{m} = 49.05 \, \text{kPa}
\]
Note: Total pressure = 49.05 kPa + 101.3 kPa = 150.35 kPa

Common Errors

1. Using gauge pressure instead of absolute pressure
2. Neglecting temperature in gas law calculations
3. Confusing density units (kg/m³ vs g/cm³)

Practice Problems

1. A hydraulic press has pistons with 10:1 area ratio. What input force lifts a 500 kg mass?
2. Calculate the moles of oxygen in a 0.02 m³ tank at 300 kPa and 293 K.
3. Explain why submarines have maximum operating depths.